Answer
$\dfrac{3+\dfrac{12}{x}}{1-\dfrac{16}{x^{2}}}=\dfrac{3x}{x-4}$
Work Step by Step
$\dfrac{3+\dfrac{12}{x}}{1-\dfrac{16}{x^{2}}}$
Evaluate the sum indicated in the numerator and the substraction indicated in the denominator:
$\dfrac{3+\dfrac{12}{x}}{1-\dfrac{16}{x^{2}}}=\dfrac{\dfrac{3x+12}{x}}{\dfrac{x^{2}-16}{x^{2}}}=...$
Evaluate the division:
$...=\dfrac{3x+12}{x}\div\dfrac{x^{2}-16}{x^{2}}=\dfrac{x^{2}(3x+12)}{x(x^{2}-16)}=...$
Take out common factor $3$ from the parentheses in the numerator and factor the denominator:
$...=\dfrac{3x^{2}(x+4)}{x(x-4)(x+4)}=...$
Simplify:
$...=\dfrac{3x^{2}}{x(x-4)}=\dfrac{3x}{x-4}$
